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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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then
Y'(t) = eat [u'(t) + au (t)]
and
Y"(t) = eat [u"(t) + 2au'(t) + a2u (t)].
Substituting for y, y, and in Eq. (27), canceling the factor eat, and collecting terms, we obtain
au"(t) + (2aa + b)u'(t) + (aa2 + ba + c)u (t) = Pn (t). (28)
The determination of a particular solution ofEq. (28) is precisely the same problem, except for the names of the constants, as solving Eq. (24). Therefore, if aa2 + ba + c is not zero, we assume that u(t) = A0tn + ... + An; hence a particular solution of Eq. (27) is of the form
Y (t) = eat (A0tn + A1tn-1 + ... + An). (29)
On the other hand, if aa2 + ba + c is zero, but 2aa + b is not, we must take u(t) to be of the form t(A0tn + ... + An). The corresponding form for Y(t) is t times the expression on the right side ofEq. (29). Note that if aa2 + ba + c is zero, then eat is a solution of the homogeneous equation. If both aa2 + ba + c and 2aa + b are zero (and this implies that both eat and teat are solutions of the homogeneous equation), then the correct form for u(t) is t2( A0tn + ... + An). Hence Y(t) is t2 times the expression on the right side ofEq. (29).
g(t) = eatPn(t) cos t or eatPn(t) sin t. These two cases are similar, so we consider only the latter. We can reduce this problem to the preceding one by noting that, as a consequence of Eulers formula, sin t = (eet - e~et )/2i. Hence g(t) is ofthe form
e(a+ie)t e(a-ie)t
g(t) = Pn (t)-----------2i--------
and we should choose
Y(t) = e(a+ie)t (A0tn + ... + An) + e(a-ie)t (B0tn + ... + Bn), or equivalently,
Y(t) = eat(A0tn + ... + An) cos t + eat(B0tn + ... + Bn) sin t.
Usually, the latter form is preferred. If a satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by t to increase their degrees by one.
If the nonhomogeneous function involves both cos t and sin t, it is usually convenient to treat these terms together, since each one individually may give rise to the
178
Chapter 3. Second Order Linear Equations
same form for a particular solution. For example, if g(t) = t sin t + 2cos t, the form for Y(t) would be
Y(t) = (A0t + A.) sin t + (B0t + Bx) cos t, provided that sin t and cos t are not solutions of the homogeneous equation.
PROBLEMS
In each of Problems 1 through 12 find the general solution of the given differential equation.
1. /- 2/- 3y = 3e^
3. /- 2/- 3y = 3te-t
5. / + 9y = t2e3t + 6 7. 2/' + 3/ + y = t2 + 3 sin t
9. u" + o^u = cos ot, 2 = 2
11.
12.
' + y + 4 y = 2 sinh t y" - - 2y = cosh2t
2. + 2 + 5 y = 3 sin 2t
4. ' + 2 = 3 + 4sin2t
6. ' + 2 y + y = 2e-t
8. ' + y = 3 sin2t + t cos2t
10. 11 + o'2 u = cos o0t
(e - - ' )/2
( + -t )/2
In each of Problems 13 through 18 find the solution of the given initial value problem.
13. / + /- 2y = 2t, y(0) = 0, (0) = 1
14. y" + 4y = t2 + 3e', y(0) = 0, (0) = 2
15. /- 2/ + y = tet + 4, y(0) = 1, (0) = 1
16. - 2/- 3 y = 3te2t, y(0) = 1, (0) = 0
17. + 4y = 3 sin2t, y(0) = 2, /(0) = -1
18. /' + 2/ + 5y = 4e-tcos2t, y(0) = 1, /(0) = 0
In each of Problems 19 through 26:
(a) Determine a suitable form for Y(t) if the method of undetermined coefficients is to be used.
(b) Use a computer algebra system to find a particular solution of the given equation.
>
>
>
>
>
>
>
19.
21.
22. y
23. /
24. /
25. y
26.
+ 3y = 2t4 + t2e-3t + sin3t > 20. + y = t(1 + sint)
- 5y + 6y = e cos2t + e2(3t + 4) sin t
+ 2y + 2y = 3e-t + 2e-t cos t + 4e-t^ sin t
- 4 + 4y = 2^ + 4te2t + tsin2t + 4y = t2 sin2t + (6t + 7) cos2t
+ 3y + 2y = '(^ + 1) sin2t + 3e-t cos t + 4e
+ 2 + 5y = 3te-t cos2t - 2te-2t cos t
27. Determine the general solution of
+ 2 y = J]
sin mn t,
m=1
where > 0 and = mn for m = 1N.
> 28. In many physical problems the nonhomogeneous term may be specified by different formulas in different time periods. As an example, determine the solution y = () of
/ + y =
t,
-t,
0 < t < , t > ,
satisfying the initial conditions y(0) = 0 and (0) = 1. Assume that y and y are also continuous at t = . Plot the nonhomogeneous term and the solution as functions of time. Hint: First solve the initial value problem for t < ; then solve for t > , determining the constants in the latter solution from the continuity conditions at t = .
3.7 Variation of Parameters
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